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Department of Electronics & Communication Engineering Laboratory Manual Analog Electronic Circuits EXPERIMENT NO.1 Design and measure the frequency response of an RC coupled amplifier using discrete components. Aim: To study the frequency response of a single stage RC coupled amplifier in a CE configuration, at low, middle, high frequency and demonstrate that gain bandwidth product is constant. Apparatus: Bread board, NPN-transistor-1 (BC 547), resistances (3.3kohm-1, 33kohm, 1kohm-1, 330 ohm, variable resistance 20kohm) capacitors (22μf, 0.1 μf, 100 μf), regulated power supply (+12), function generator, C.R.O., connecting wires. Theory: A circuit diagram of RC coupled amplifier in CE configuration is shown in fig-1. The gain of an amplifier depends on frequency .As frequency decreases the gain starts to fall .The main cause of this is coupling capacitor Cc. Because it offers very high impedance at low frequency this causes decrease in output voltage. At high frequency the gain again falls because of the shunt capacitances made up of junction capacitances and wiring capacitances. The size of the coupling capacitor is so chosen that it offers negligible reactance to ac at operating frequencies. Ce acts as a by pass capacitor for ac signals.Cb is that it must pass the input ac signal unattenuated. Hence the lower cut off frequency of the amplifier is determined by the coupling capacitors. The Coupling capacitor Cb along with resistance combination forms a high pass filter whose cut off frequency, fl = 1 / 2πRC Shunt capacitance at the output and RL + hi acts as low pass filter (high frequency response). The gain of the amplifier at different frequency is: At middle frequency The current gain is hfe R L Ai = hie + R L The Voltage gain is hfe R L Av = hie + R L At low frequency hfe R L the current gain is Ai = hie + R L - J C The Voltage gain is hfe R L Av = hie + R L - J C At high frequency hfe R L the current gain is Ai = hie + R L + J Cd R L hie The Voltage gain is hfe R L Av = hie + R L + J Cd R L hie where hie is the input impedance R L is the load impedance Cd is the shunt capacitance(junction + wiring capacitance) The lower cut off frequency at which gain the gain falls to 70.7% of its maximum value and the higher cut off frequency for which the gain falls to 70.7% of its maximum value. Bandwidth of an amplifier is given by BW = fh-fl This type of amplifier generally used in audio frequency range. The frequency response curve of this amplifier is shown in fig-2. Fig.-1 Circuit diagram of single stage RC Coupled Amplifier Where Vcc = +12v R1 33kΩ , R2 3.3kΩ , R3 1kΩ, R4 Variable 20kΩ Cb 0.1μf Ce 22μf Cc 100μf Gain BW = gain x (fh-fl) Fig-2 Frequency response of RC-coupled Amplifier The gain of RC coupled amplifier will fall at the rate of 20 db per decade from the unity gain bandwidth. Procedure: 1. Make connection as shown in fig.1. 2. After making sure that transistor is biased in active region feed the input signal from function generator (sine wave amplitude of 1mv) such that the output signal is undistorted. 3. Vary the input signal frequency from the function generator and observe for the voltage output (Vo) at the CRO, adjust the pot meter R4 to get the overall gain of 10. 4. Find the maximum voltage output by varying the frequency of input signal. Make sure that the amplitude of the input signal should remain same. 5. Record the voltage output in observation table by decreasing the frequency in small steps, below the frequency at which maximum voltage is obtained. Take the reading till voltage output is readable on the CRO. 6. In the similar manner, find the higher cut off frequency (fh) by increasing the frequency from the frequency where maximum gain is obtained. 7. Record the voltage output in observation table by increasing the frequency in small steps, above the frequency at which maximum voltage is obtained. Take the reading till voltage output is readable on the CRO. 9. Plot the frequency response curve as per the reading noted in the observation table. 10. Find out the lower cut-off frequency and higher cut-off frequency, where the gain falls up to 70.7% of maximum gain. 10. Calculate the bandwidth (f’h – f’l). 11. Repeat the steps from step no.3 to 10, for the gain of 20 by adjusting the R4 and make sure that amplitude of input signal should not change. 13. Find the gain bandwidth product and verify that this product is constant. Observations Table: S.No. Frequency of input signal Output Voltage Gain (10) For single stage: Higher cut off frequency (fh)=…………. Lower cut off frequency (fl)=………… BW = fh - fl Precautions: 1. Make connection very carefully. 2. Make connection firm & tight 3. Use C.R.O. carefully. Gain (20) EXPERIMENT NO.2 Design a two stage RC coupled amplifier and determine the effect of cascading on gain and bandwidth. Aim: To study the frequency response of two stages RC coupled amplifier at low, middle, high frequency and demonstrate that the gain bandwidth product is constant. Apparatus: Breadboard, NPN-transistor (BC 547), resistances, capacitors, regulated power supply (+12), function generator, C.R.O., connecting wires. Theory: A circuit diagram of two stages RC coupled amplifier is shown in fig-1. The gain of two stage amplifier depends on frequency .As frequency decreases the gain starts to fall .The main cause of this is coupling capacitor Cc. Because it offers very high impedance at low frequency this causes decrease in output voltage. At high frequency the gain again falls because of the shunt capacitances made up of junction capacitances and wiring capacitances. The Higher cut off frequency for which the gain falls to 70.7% of its maximum value is given by: f’h = fh (21/n -1 )1/ 2 The lower cut off frequency at which gain the gain falls to 70.7% of its maximum value is given by: f’L = fl i n 2 1 Bandwidth of two stage amplifier is given by: BW = f’h - f’L Gain BW = gain x (fh-fl) Where fh is single stage higher cut off frequency fL is single stage lower cut off frequency n is no. of single stage in cascading f’h is multistage higher cut off frequency f’L is multistage lower cut off frequency Fig.-1 Circuit diagram of two stage RC coupled amplifier Where Vcc = +12v R1 33kΩ , R2 3.3kΩ , R3 1kΩ, R4 Variable 20kΩ Cb 0.1μf Ce 22μf Cc 100μf fig. 2 Frequency response of RC-coupled Amplifier Procedure: 1. Make connection as shown in fig.1. 2. After making sure that transistors are biased in active region feed the input signal from function generator (sine wave amplitude of 1mv) such that the output signal is undistorted. 3. Vary the input signal frequency from the function generator and observe for the voltage output (Vo) adjust the pot meter R4 to get the overall gain of 10. 4. To observe the frequency response of the first stage disconnects the second stage by removing the right lead of Cc, which is connected, to the base of second transistor. 5. Find the maximum voltage output by varying the frequency of input signal. 6. Record the voltage output in observation table by decreasing the frequency in small steps, below the frequency at which maximum voltage is obtained. Take the reading till voltage output is readable on the CRO. 7. In the similar manner, find the higher cut off frequency (fh) by increasing the frequency from the frequency where maximum gain is obtained. 8. Record the voltage output in observation table by increasing the frequency in small steps, above the frequency at which maximum voltage is obtained. Take the reading till voltage output is readable on the CRO. 9. Plot the frequency response curve as per the reading noted in the observation table. 10. Find out the lower cut-off frequency and higher cut-off frequency from the graph, where the gain falls up to 70.7% of maximum gain. 11. Calculate the bandwidth (f’h – f’l). 12. Now to observe the effect of cascading, now before connecting the second stage with first stage, make sure that second stage of amplifier is identical and its gain as a single amplifier is same as stage first i.e.10 by adjusting the R4. 13 After making sure that transistors are biased in active region feed the input signal from function generator (sine wave amplitude of 1mv) such that the output signal is undistorted. 14 Find the maximum voltage output by varying the frequency of input signal. 15. Repeat the steps from step no.6 to 10. 16. Calculate the bandwidth of two stage amplifier as BW = f’h - f’L 17. Find the gain bandwidth product for both stages. Gain BW = gain x (fh-fl) 18. Verify the gain and bandwidth product is constant Observations Table: For single stage: S.No. Frequency of input signal Output Voltage Gain (10) For double stage: S.No. Frequency of input signal For single stage: Higher cut off frequency (fh)=…………. Lower cut off frequency (fl)=………… BW = fh - fL For double stage: Higher cut off frequency (f’h)=…………. Lower cut off frequency (f’l)=………… BW = f’h - f’L Output Voltage Precautions: 1.Make connection very carefully. 2. Make connection right & tight 3. Use C.R.O. carefully. 4. Draw curve carefully and it should be like fig.2 EXPERIMENT NO. 3 Design of RC phase shift oscillator using discrete components. Aim: To design and realize RC phase shift oscillator of having frequency of 200 hz using discrete components. Apparatus: BC-547, resistances (10k ohm-1, 47k ohm-1, 3.3kohm-3, capacitors (0.1μf -3), regulated power supply (+12v), CRO, connecting wires, breadboard. Theory: The circuit arrangement of a Phase –shift oscillator using N-P-N transistor in CE configuration is shown in fig-1. As usual, the voltage divider R1- R2 provides dc ec emitter base bias, Re andCe combination provides temperature stability and prevent ac signal degeneration and collector resistor Rc controls the collector voltage. The oscillator output voltage is capacitevely coupled to the load by Cc.In case of a transistor phase- shift oscillator, the output of the feedback network is loaded appreciably by the relatively small input resistance (hie) of the transistor. Hence instead of employing voltage series feedback, voltage shunt feedback is used for a transistor phase shift oscillator. The output of the amplifier is fed back to the input through RC network. This RC network provides an overall phase shift of 1800 .This phase shift is achieved in three stages of RC network. The additional 1800 phase shift is obtained from transistor itself. Thus, the phase of the output is in same phase with the input. Output waveform of RC phase shift oscillator is shown in fig-2. It can be shown that frequency at which RC network provides exactly 1800 phase shift is given by fo = 1/ 2π √6 +4RC /R If R=Rc Then fo = 1/ 2π RC√10 When the feedback is, the overall gain of the amplifier is written as Af = A/ (1-Aβ), Where Aβ is feedback factor or loop gain. If Aβ =1 , Af =∞. Thus the gain becomes infinity i.e. there is output without any input. In another words, the amplifier works as oscillator. This condition is known as Barkhausen criterion of oscillation. For the loop gain to be greater than unity, the requirement of the current gain of the transistor is found to be hfe > 23+29 (R/RC) + (Rc / R) If R=Rc hfe > (23+29 + 4) i.e. >56 The phase shift oscillator is well suited to the range of frequencies from 20 Hz to 200kHz, and so includes the audio frequency range up to 20kHz. Fig.1 Circuit diagram of Phase shift oscillator R1=33kohm R2 =3.3kohm Rc=1kohm Ce= 22microfarad Re=variable resistance (5kohm) C=0.1microfarad R=3.3 kohm Procedure: 1. Connect the circuit as shown in fig-1. 2. Adjust the pot-meter for the gain of 56 or more to get the oscillations ,observe the output from collector on the C.R.O. 3. Note this output will be 180o out of phase with respect to input at base of the transistor. 4. Measure the frequency of voltage out. 5. Calculate the frequency of oscillation from formula and compared with the observed value. Fig-2 Phase shift oscillator out waveform Precautions: 1. All the connection must be tight. 2. Amplifier gain should be more than 56 to sustain the oscillations. EXPERIMENT NO. 4 Design and realize inverting amplifier, non - inverting amplifier and buffer amplifier using 741 Op Amp. Aim: To design inverting amplifier, non –inverting amplifier for the gain of 10 and establishing the saturation input voltage level and design the dc buffer. Apparatus: Op-amp 741, two resistors (1kΩ), one resistor (10kΩ), bread board, connecting wires, CRO, function generator, multimeter, regulated power supply (+12v & -12v), variable power supply (0-30v). a) Inverting amplifier: Theory: A circuit diagram of an inverting amplifier using operational amplifier as shown in fig 1 (a) and fig -1(b) shows the pin diagram of op-741and fig-2, fig-3 shows input and output waveforms of an inverting amplifier. The input signal is applied to the inverting terminal (pin-2) of the operational amplifier 741. Non-inverting terminal (pin-3) is connected to the ground. As these two terminals 2 and 3 are at virtual ground; the potential at point p will nearly be equal to zero. Apply KCL at point P. The current “ i1 flowing through point p is given by i1 = (v1 – vi ) / R1 ( p is at ground potential) i1 = v1 / R1 Similarly, i2 = (vi – v0 ) / Rf i2 = - v0 / Rf At point p v1 / R1 = - v0 / Rf The ratio of output voltage v0 and input voltage vi is known as the gain of the amplifier. Hence the gain of inverting amplifier is given by Av = v0 / v1 = - Rf / R1 vo = -(Rf / Ri) vi The above equation shows that vo will be 1800 out of phase with respect to input. Rf i2 10 kΩ v1 R1 1kΩ 4 -12v 2 6 +ve p vo 3 i1 7 Ro +12v Fig -1(a) Circuit diagram of Inverting Amplifier Offset null 1 Inverting input 2 Non- Inverting input V- 3 4 8 Op-741 6 8 NC 7 V+ 6 Output 5 Offset null Fig-1(b) Pin diagram of op-amp 741 Fig-2(a) Input waveform of Inverting Amplifier Fig-2 (b) Output waveform of Inverting Amplifier Procedure: 1. Connect the circuit diagram as shown in fig-1(a) for inverting amplifier. 2. Connect the power supply +12v & -12v to pin no7 & 4 respectively. 3. Apply the dc input signal to pin 2 of Op- amp-741 in step of 0.25v and measure the output voltage with the help of multimeter. 4. Note down the input and output voltages in the observation table. 5. Observe the gain of Op- amp 741, which is dependent on Rf and R1. 6. Verify the gain with the theoretical values. 7. Find out the saturation input voltage level. 8. Same experiment may be repeated with the ac input signal from the function generator and observe the waveforms on the CRO. Sl.no Input voltage Output voltage Gain ( Av = v0 / v1) b) Non - inverting amplifier: Theory: The circuit configuration of an op-amp working in the non-inverting mode is shown in the fig-3 and fig-4 shows the input and output waveforms of a non-inverting amplifier. Here the input voltage v2 is applied to the non – inverting terminal (pin-3) of operational amplifier and hence the name is non-inverting amplifier. The potential of point p is also v2 since the gain of op-amp is infinite. The polarity of vo is the same as that of v2. The voltage across R1 is v2 and across Rf is (vo – v2). The values of currents i1 and i2 are given as i1 = v2 / R1 i2 = (vo – v2 ) / Rf Applying Kirchhoff’s first law at point p, we get (- i1) + i2 = 0 - v2 / R1 + (vo – v2 ) / Rf = 0 vo / v2 = (1 + Rf / R1) Av = (1 + Rf / R1) vo = (1+Rf / Ri) vi This is shown that properly selecting the values of resistance Rf and R1 can control the output voltage. i2 –12v R1 p 1 kΩ 4 2 Rf 10kΩ +ve 6 3 i1 v2 7 +12v Fig-3 Circuit diagram of Non- Inverting Amplifier Vo Fig-4 (a) Input waveform of Non- Inverting Amplifier Fig-4 (b) Output waveform of Non- Inverting Amplifier Procedure: 1. Connect the circuit diagram as shown in fig-3 for non- inverting amplifier. 2. Connect the power supply +12v & -12v to pin no7 & 4 respectively. 3. Apply the dc input signal to pin 3 of Op- amp-741 in step of 0.25v and measure the output voltage with the help of multimeter. 4. Note down the input and output voltages in the observation table. 5. Observe the gain of Op- amp 741, which is dependent on Rf and R1. 6. Verify the gain with the theoretical values. 7. Find out the saturation input voltage level. 8. Same experiment may be repeated with the ac input signal from the function generator and observe the waveforms on the CRO. Observation Table: Sl.no Input voltage Output voltage Gain Av = (1 + Rf / R1) C) Buffer amplifier: Theory: A circuit diagram of buffer or dc voltage follower is shown in fig-5. If R1 is infinite, then Av =1 i.e. vo = vi. Thus the output voltage follows the input voltage. It means this circuit is acting as the voltage follower. The input & output are tied together. Such circuit has a very high input resistance (10,000 M ohm), and very low output resistance., this is generally used as impedance matching device or buffer between high impendence source and low impedance load. -12v 4 2 6 vo Vi 10 kΩ 3 7 +12v Fig-5 Circuit diagram of DC voltage follower Sl.no Input voltage Output voltage Gain ( Av = v0 / v1) Procedure: 1. Connect the circuit on the breadboard as shown in fig-1. 2. Apply input at pin 3 from the dc source. 3 Note down the input and output voltages in the observation table with the help of multimeter. 4. Verify that input voltage is equal to output voltage. Precautions: 1. The CRO should be handled carefully. 2. Check for proper connections. EXPERIMENT NO.5 Verify the operation of differentiator circuit using op-amp 741and that it acts as a high pass filter. Aim: Designing a differentiator which is acting as high pass filter having a cut-off frequency of 1khz Apparatus: Bread Board, connecting wires, resistances (1k, 10k ohm), capacitor (0.1 f ),function generator, CRO, probes, regulated power supply (+12v & -12v). Theory: The circuit performs the mathematical operation of differentiation i.e. the output waveform is the derivative of the input waveform. It is constructed from a basic inverting amplifier if an input resistor R1 is replaced by capacitor C. Vo = - (RC) dVi/dt The output voltage is the differentiation of input voltage. The circuit diagram of differentiator using op-741 is shown as fig-1 and input / output waveforms shown as fig-2. When we feed linearly increasing voltage to the differentiator, we get a constant dc output. So it is an inverse mathematical operation to that of an integrator. Assuming that p is at ground potential, we can write for capacitor v1 = q /C Where q is charge on capacitor d v1 / dt = 1 / C (dq /dt ) = i /C , Where i = dq /dt Again, vo = -RC . dv1/dt Shows that the output voltage vo is equal to a constant –RC times the derivative of input voltage v1. The differentiator is most commonly used in wave shaping circuits to detect high frequency components in an input signal and also as a rate of change detector in FM modulator. The frequency response of the differentiator is shown in fig-3. In this figure ,fa is the frequency at which the gain is 0 db and is given by fa = 1 / (2πRC) Also, fc is the unity gain bandwidth of the op-amp. fb is the gain limiting frequency . R 1.5kΩ C v1 0.1f i -12v 4 p 2 6 Vo 3 7 + 12v Fig-1 Circuit diagram of op-amp 741 as differentiator Vi Fig-2(a). Input waveform of differentiator Vo Fig-2(b). Output waveform of differentiator Open-loop Gain differentiator closed- loop response of basic fb fa closed loop response of practical differentiator Frequency fc Fig-3 Frequency response of basic differentiator circuit Procedure: 1. Connect the circuit on the breadboard as shown in fig-1. 2. Apply input at pin 2 from the function generator. 3. Observe the input and output waveforms on the CRO as shown in fig 2 &3. 4. Vary the frequency of the input signal and note the amplitude of the output along with input frequency. 5. Plot the graph between output voltage and the frequency of the input signal. 6 Verify the frequency response of circuit with the theoretical value i.e. 1khz in this case. Observation Table: Sr. No. 1. 2. 3. 4. 5. 6. 7. f(Hz) Vo Precautions: 1.The CRO should be handled carefully. 2 Check for proper connections. EXPERIMENT NO. 6 Verify the operation of an integrator circuit using op-amp 741and show that it acts as a low pass filter. Aim: Designing an integrator, which is acting as low pass filter having a cut-off frequency of 1 kHz. Apparatus: Bread board, connecting wires, resistances (1k, 1.5k, 10k ohm), capacitor (0.1 f), function generator, CRO, probes, regulated power supply (+12v & -12v). Theory: A circuit in which the output voltage waveform is the integral of the input voltage waveform is known as integrating amplifier. Such a circuit is obtained by using a basic inverting amplifier configuration if the feed back resistor Rf is replaced by capacitor C.Feedback through the capacitor forces a virtual ground to exist at the inverting terminal. Now the voltage across the condenser is simply the output voltage vo. The capacitive impedance Xc can be expressed as Xc = 1 / jω C = 1 / s C Where s = j ω (Laplace notation) From the fig-1, i1 = v1 /R1 and i2 = - vo / Xc = - s C vo At point p , i1 = i2 v1 /R1 = - s C vo vo /v1 = - (I / s C R1) This equation can be written in time domain as vo (t) = - (1/R1 C) ∫ v1 dt Fig-1 indicates that the output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RC. If the input is a sine wave, the output will be a cosine wave and if the input is a square wave the output will be a triangular wave as shown in fig-2 & 3. The circuit diagram shown in fig-1 can also acts as low pass filter. At low frequencies, the capacitor appears as open . In this case the circuit acts like an inverting amplifier. The voltage gain is (-R /R1). As the frequency increases, the capacitive reactance decreases. Now the voltage gain drops. When the frequency approaches infinity, the capacitor appears shorted . So the voltage gain approaches zero. Fig -4 shows the output response. Following relation will calculate the cutoff frequency, fc = 1 / (2πRC) To design a low pas filter with cutoff frequency of 1khz, the values of R & C may be calculated. The values we get is R= 1.5kΩ and C = 0.1f. 0.1f C R 1.5kΩ R1 1 kΩ vi -12v p 4 2 6 vo 3 7 +12v Fig-1 Circuit diagram of an Integrator using op-amp 741 Vi t Fig-2 (a) Input waveform of an Integrator Vo t Fig-2(b) Output waveform of an Integrator Fig-3 (a) Input waveform of an Integrator Fig-3(b) Output waveform of an Integrator Fig-4 Output response Procedure: 1. Connect the circuit on the breadboard as shown in fig-1. 2. Apply input at pin 2 from the function generator. 3. Observe the input and output waveforms on the CRO as shown in fig 2 &3. 4. Vary the frequency of the input signal and note the amplitude of the output along with input frequency. 5. Plot the graph between output voltage and the frequency of the input signal. 6 Verify the frequency response of circuit with the theoretical value i.e. 1khz in this case. Observation Table: Sr. No. 1. 2. 3. 4. 5. 6. 7. f(Hz) Vo Precautions: 1. The CRO should be handled carefully. 2.Check for proper connections. EXPERIMENT NO. 7 Design and verify the operation of adder and subtractor circuit. Aim: To design and realize the adder with the gain of 10 and subtractor circuit using Op- Amp 741. Apparatus: Op-amp 741, 4 resistors (1kΩ), 3 resistors (10kΩ), 1 resistor (5kΩ) bread board, connecting wires, variable power supply 0-30v, regulated power supply(+12v & 12v), multimeter. Theory: a) Operational amplifier 741 as adder The circuit that performs the addition of three signals with amplification (with gain of 10) is shown in fig1. The most useful of the op-amp741 circuits used in analog computers is the summing amplifier circuit. This circuit provides a means of algebraically summing three input voltages, each multiplied by a constant gain factor. In the circuit p is virtual ground and the output voltage is phase inverted. As p is virtual ground, the different currents are given by i1 = v1 / R1 , i2 = v2 / R2, i3 = v3 / R3 and i = - vo / Rf Applying Kirchhoffs current distribution law at point p , we get i1 + i2 + i3 - i = 0 or v1 / R1+ v2 / R2 + v3 / R3 - vo / Rf =0 vo = - {( Rf / R1)v1 + (Rf / R2)v2 +( Rf / R3)v3 If = R3 =R and Rf / R = k, then vo = - k { v1 + v2 + v3 } vo = - Rf / R { v1 + v2 + v3} Thus the output voltage is proportional to the algebraic sum of three input voltages. Again if Rf = R, Note that the input signals can be added together with different weighing factors, if desired. However, if all the resistors are chosen to be equal, the circuit acts as a pure adder and adds the input voltages together at the output i.e. in case R1 = R2 = R3 , the circuit acts as a pure adder and output voltage is given by: vo = -{ v1 + v2 + v3 } Rf R1 v1 1kΩ i1 10kΩ 2 R2 1kΩ v2 i2 4 -12v p 6 v3 1kΩ i3 3 7 +12v R13 Fig –1 Circuit diagram of an adder Procedure: 1.Connect the various components according to the fig -1 shown. 2.Apply the input voltages from the dc source and increase in the steps . 3. Measure the input and output voltages with the help of multimeter and record the reading in the observation table . 4. Verify the output voltage with the theoretical value. Observation Table Sl.no Input voltage Output voltage Theoretical values vo = - Rf / R { v1 + v2 + v3} Vo b) Op-Amp as a subtractor: Theory: Since differential amplifiers amplify the difference of two input signals applied to the inverting and non-inverting terminals of an op-amp, it can be used as subtractor circuit. The circuit is shown in fig-2. This circuit provides an output equal to the difference of two input signals. In this circuit one signal is applied at the inverting terminal while the second signal is applied at non-inverting terminals of an op-amp. Let (vo)1 and (vo)2 be the output voltages produced by input voltages v1 and v2 respectively. (vo)1 = -( Rf / R1) v1 (vo)2 = (1+Rf / R1) v2 = (Rf / R1) v2 (Rf >> R1) According to superposition theorem vo = (vo)1 +(vo)2 vo = -( Rf / R1) v1 + (Rf / R1) v2 vo = Rf / R1 (v2 - v1) . Thus the output voltage vo is equal to the voltage applied to the non-inverting input terminal less the voltage to the inverting input terminal. Hence the circuit is called a subtractor. Rf v1 v2 R1 1kΩ R2 1kΩ -12v 22kΩ p 4 2 6 3 7 +12v Fig-2 Circuit diagram of subtractor Procedure: vo 1. Connect the various components according to the fig -1 shown. 2. Apply the input voltages from the dc source and increase in the steps. 3. Measure the input and output voltages with the help of multimeter and record the reading in the observation table. 4. Verify the output voltage with the theoretical value. Observation Table Sl.no Input voltage Output voltage V0 V1 V2 Precautions: 1. The CRO should be handled carefully. 2. Check for proper connections. Theoretical values vo = Rf / R1 (v2 v1) EXPERIMENT NO. 8 To design and realize using op-amp741, Wein- bridge oscillator. Aim: To design and realize Wein- bridge oscillator using discrete components with opamp741 having frequency of approximately 1khz (965hz). Apparatus: Op-amp741, resistances (3.3k ohm-2, 12k ohm-1,33k ohm-1, capacitors (0.05microfarad-3), regulated power supply(+12v &-12v) , CRO, connecting wires, breadboard. Theory: Fig-1 shows the circuit diagram of Wein- bridge oscillator. In this circuit the Wienbridge circuit is connected between the amplifier input terminal and the output terminal. The bridge has a series RC network in one arm and a parallel RC network in the adjoining arm. In the remaining arms of the bridge resistors R1 and Rf are connected. The phase angle criterion for oscillation is that the total phase shift around the circuit must be 0 degree. The condition occurs only when the bridge is balanced, that is at resonance. The frequency of oscillation fo is exactly the resonant frequency of the balanced Wien-bridge and is given by fo = 1 / 2 πRC or fo = 0.159 / RC Assuming that the resistors are equal in value, and the capacitors are equal in value in the reactive leg of the wein-bridge.At this frequency the gain required for sustained oscillation is given by Av = 1 / B = 3 That is , 1+ Rf / R1 =3 or Rf = 2R1 R 12k -12v 33k k 2 3 4 Vout 6 7 +12v R C 0.05f 3k3 R 3 k 3 C 0.05f Fig.1 Circuit diagram of Wein- bridge oscillator Procedure: 1. Connect the circuit as shown in fig-1. 2. Take the output from pin 6 of the op-amp and observe it on CRO. 3. Note the frequency of the output and compare with the theoretical value, which may be calculated by the above equation.. Precautions: 1.All the connections must be tight. 2.Op-amp should be connected properly. EXPERIMENT NO. 9 To design and realize using op-amp741, square wave generator.. Aim: To design and realize square wave generator using discrete components with opAmp741 having frequency of 1khz. Apparatus: Op-amp741, resistances (10k ohm-4, 47k ohm-1, capacitors (68nf-3), regulated power supply (+12v &-12v), CRO, connecting wires, breadboard. Theory: The circuit is uses a comparator with both positive and negative feedback to control its output voltage. Because the negative feedback path uses a capacitor while the positive feedback path does not, however, there is a time delay before the comparator is triggered to change state. As a result, the circuit oscillates, or keeps changing state back and forth at a predictable rate. Because no effort is made to limit the output voltage, it will switch from one extreme to the other. If we assume it starts at -10 volts, then the voltage at the "+" input will be set by R2 and R1 to a fixed voltage equal to -10R1/(R1 + R2) volts. This then becomes the reference voltage for the comparator, and the output will remain unchanged until the "-" input becomes more negative than this value. But the negative input is connected to a capacitor (C) which is gradually charging in a negative direction through resistor Rf. Since C is charging towards -10 volts, but the reference voltage at the positive input is necessarily smaller than the 10 volt limit, eventually the capacitor will charge to a voltage that exceeds the reference voltage. When that happens, the circuit will immediately change state. The output will become +10 volts and the reference voltage will abruptly become positive rather than negative. Now the capacitor will charge towards +10 volts, and the other half of the cycle will take place. With the op-amp output voltage at negative saturation, -Vs at, the across R1 is also negative, since The voltage at v1 = R1 / R1 + R2 (-Vsat) With output at +Vsat, voltage v1 at the non-inverting input is v1 = R1 / R1 + R2 (+Vsat) The output frequency is given by the equation below: fo = 1/ {2 R C ( 2R1 + R2 / R2 )} . R 0.05f -12v 10k v2 k 2 3 C 4 7 +12v v1 1 0 k Vout 6 2 0 k R2 R1 Fig.1 Circuit diagram of Square wave generator Procedure: 1. Connect the circuit as shown in fig-1. 2. Take the output from pin 6 of the op-amp and observe it an CRO. 3. Note the frequency of the output and compare with the theoretical value, which may be calculated by the above equation.. Precautions: 1.All the connections must be tight. 2.Op-amp should be connected properly. EXPERIMENT NO-10 Design and realize AC voltage follower using op-amp 741. Aim: Design and verify the operation of AC voltage follower using ac input signal from the function generator. Apparatus: Breadboard, op-amp 741, connecting wires, CRO, resistances (10 k, 100kΩ), capacitors .01μf and 2μf., regulated power supply (+12v & -12v), function generator. Theory: AC Voltage follower A circuit diagram of AC voltage follower is shown as fig-1 This circuit is used to provide impedance buffering i.e., to connect a signal source with high internal source impedance to a load of low impedance, which may even be capacitive. Fig-1 shows a practical high input impedance ac voltage follower making use of op-amp. We assume that C1 & C2 present short circuit at all frequencies of operation for this circuit. Resistance R1 and R2 are used to provide RC coupling and allow a path for the dc input current into the non-inverting terminal. In the absence of capacitor C2, the ac signal source would see an input resistance of only R1 + R2 = 200 kohm. Since the op amp is connected as a voltage follower, the voltage gain Av between the output terminal and the non-inverting terminal is very close to unity. -12v 4 0.01μ f C1 2 3 6 Vo Vi 7 +12v R1 100kΩ C2 2μf R2 100kΩ Fig-1 Circuit diagram of AC voltage follower Procedure: 1. Connect the circuit on the breadboard as shown in fig-1. 2.Apply the input signal from the function generator at pin 3. 3.Observe the input and output waveforms on the CRO. 4. Vary the amplitude of the input signal and note the input and output amplitude. Observation Table: Sr. No. 1. 2. 3. 4. 5. 6. 7. Input Output Voltage Voltage Precautions: 1. The CRO should be handled carefully. 2. Check for proper connections. EXPERIMENT NO. 11 Design and realize op-amp-741 as a voltage comparator. Aim: Design and verify the operation of op-amp-741 as a voltage comparator by showing the output using LEDs. Apparatus: Breadboard, op-amp 741, LEDS-2, resistors (1 eke)-2, regulated power supply (+5v,+12v & -12v), variable dc power supply (0- 30 v). Theory: Comparator is a circuit that compares a signal voltage applied at the input of an op-amp with a known reference voltage given at the other input. Basically comparator is an open-loop operation i.e., there is no feedback path in the case of comparator. Fig-1 shows the schematic circuit of comparator. -12v (0-30 v ) 4 2 Vo Vi 741 6 3 7 Vr 1KΩ +5v 1KΩ +12v D1 D2 Fig-1 Circuit diagram of voltage comparator The input voltage, Vi, to be compared with reference voltage is applied to the inverting terminal. The reference voltage is applied to the non-inverting terminal of the op- amp. i) ii) iii) Vi < Vref : Vi > Vref : Vi = Vref: Output, Vo is positive; LED D1 will glow. Output, Vo is negative; LED, D2 will glow. No LED will glow. Procedure: 1. Connect the circuit on the breadboard as shown in fig-1 above. 2.Apply varying input voltage to the inverting terminal (pin-2) of the op-amp. 3.Apply fixed reference voltage (say 5V) to the non-inverting terminal of op- amp. 4. Measure the input and output voltages with the help of multimeter and record in the observation table. Observation Table: Sr. No. Input voltage (Vi ) Which LEDs glowing 1 2 3 4 5 6 7 Output voltage (vo) Vi < Vref Vi = Vref Vi > Vref Plot the curve between input and output voltages. It will show clearly how the output switches from positive to negative value. Precautions: 1. The CRO should be handled carefully. 2. Check for the proper connections.